# Write a program to test various operations on a binary search tree

If the given key is equal to the key at the root, we return the number of keys t in the left subtree; if the given key is less than the key at the root, we return the rank of the key in the left subtree; and if the given key is larger than the key at the root, we return t plus one to count the key at the root plus the rank of the key in the right subtree.

Why not use the predecessor? Add to BST. In-order traversal: In this traversal mode, one starts visiting with the left child, followed by its parent and then the right child.

First, we need to find the node that will replace the deleted node. If the left link of the root is null, the smallest key in a BST is the key at the root; if the left link is not null, the smallest key in the BST is the smallest key in the subtree rooted at the node referenced by the left link.

The possible substitute would be either 1 i.

## Binary search tree c++

Finding the maximum key is similar, moving to the right instead of to the left. The three most commonly used traversing methods are: In-order Pre-order Post-order 1. Here since 11 is the root we will not visit it but will move to its left child that is 2. If a given key key is less than the key at the root of a BST, then the floor of key the largest key in the BST less than or equal to key must be in the left subtree. The cases to consider for the element to be deleted are as follows: Case 1: The element to be deleted is a leaf node. We'll use the same tree that we used before and we'll show the traversal order for each case. Reverse a BST. For delete the minimum, we go left until finding a node that that has a null left link and then replace the link to that node by its right link. That again 2 is a root and it also has a left child which is 1 so first we visit 1, and move to the right child of 2. The right child here is 9, but is a root and also has a left child thus we traverse to the left child 3. In worst case, the time it takes to search an element is 0 n. The diagram above depicts the situation described as under case 1. We keep going to either right subtree or left subtree depending on the value and when we reach a point left or right subtree is null, we put the new node there. After this, we move on to the right sub-tree.

From the example quoted above, we delete the element

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